Question

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

Answer 1

Temperature of the helium atom, T_He = –20°C= 253 K

Atomic mass of argon, M_Ar = 39.9 u

Atomic mass of helium, M_He = 4.0 u

Let, (v_rms)_Ar be the rms speed of argon.

Let (v_rms)_He be the rms speed of helium.

The rms speed of argon is given by:

`(v_"rms")_"Ar" = sqrt((3RT_"Ar")/M_"Ar")` .... (1)

Where,

R is the universal gas constant

T_Ar is temperature of argon gas

The rms speed of helium is given by:

`(v_"rms")_"He" = sqrt((3RT_"He")/M_"He")` ... (ii)

It is given that

(v_rms)_Ar = (v_rms)_He

`sqrt((3RT_"Ar")/ M_"Ar")` = `sqrt((3RT_"He")/M_"He")`

`T_"Ar"/M_"Ar" = T_"He"/M_"He"`

`T_"Ar" = T_"He"/M_"He"  xx M_"Ar"`

`= 253/4 xx 39.9`

= 2523.675 = 2.52 × 10³ K

Therefore, the temperature of the argon atom is 2.52 × 10³ K.

Answer 2

Let C and C’ be the rms velocity of argon and a helium gas atoms at temperature T K and T K respectively

Here, M = 39.9; M’ = 4.0; T =?; T = -20 + 273 = 253 K

Now, `C = sqrt((3RT)/M
)= sqrt((3RT)/39.9)` and `C' = sqrt((3RT')/M') = sqrt(3R xx 253)/4`

Since C =C'

Therefore `sqrt((3RT)/39.9) =  sqrt(3Rxx253)/4` or `T = (39.9 xx 253)/4  = 2523.7 K`